BUT WAIT A MINUTE!! why are some of the C's for the primer in the methylated version converted to T's. I thought that if you assume something is already "methylated" then shouldn't the methylated version equal exactly the wild type version? In addition why are only some of the C's converted and not all?

The original article gives a clue: stating that "the program will generate two versions ... one is the bisulfite-modified and methylated sequence in which all c's except 5mc are converted T's,... and unmethylated sequence in which all c's as well as 5mcs are converted to t's.

How in the world does the program predicte which c's are methylated base on a sequence? for example why are the 3c's in bold converted to t's in the methylated version and not the last c (bold red)?

Is there a study I am not aware off? something to do with CpG patterns?

this is of course due to my ignorance. Can anyone point me to the right direction? thank you.

apparently all non-cpg dinucleotides are unmethylated and thus need to be converted T's for the primer design. I think this is correct. The last c was actually a cpg (cut out) and thus was not converted in the methyl primer set.

Alex

-simplitia-

You are right.... If you note, the second C from the 5' end is also not converted into a T in the methylated version since it is followed by a G.